	\section{极限、连续}
	\subsection{函数极限}
	\begin{ti}
		求 $\lim_{x\to 0} \frac{\sqrt{1+x} - 1 - \frac{x}{2}}{\ee^{x^{2}}-1}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0} \frac{\ee^{x} + \ln(1 - x) - 1}{x - \arctan x}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0} \frac{(1+x)^{\frac{2}{x}} - \ee^{2}[1 - \ln(1+x)]}{x}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0} \frac{\left(1 + x^{2}\right)(1 - \cos 2x) - 2x^{2}}{x^{4}}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0} \frac{\sqrt{1-x^{2}} \sin^{2}x - \tan^{2}x }{x^{2}[\ln(1+x)]^{2}}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0} \frac{(3 + 2 \tan x)^{x} - 3^{x}}{3 \sin^{2}x + x^{3} \cos\frac{1}{x}}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 2}\frac{\sqrt{5x - 1} - \sqrt{2x + 5}}{x^{2} - 4}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0}\int_{0}^{x} \frac{\sin 2t}{\sqrt{4+t^{2}}\int_{0}^{x} \left(\sqrt{t+1} - 1\right)\dd{t}} \dd{t}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to \infty} \ee^{-x} \left( 1 + \frac{1}{x} \right)^{x^{2}}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 3^{+}} \frac{\cos x \ln (x - 3)}{\ln\left( \ee^{x} - \ee^{3} \right)}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to \infty} x^{2} \left( a^{\frac{1}{x}} + a^{-\frac{1}{x}} - 2 \right)$，其中常数 $a > 0$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0}\frac{1}{x}\left( \cot x - \frac{1}{x} \right)$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to +\infty}\left( \sqrt[3]{x^{3} + 2x^{2} + 1} - x\ee^{\frac{1}{x}} \right)$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0}\left( \frac{1+x}{1-e^{-x}} - \frac{1}{x} \right)$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0^{+}} x^{\ln\left( \frac{\ln x - 1}{\ln x + 1} \right)}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to \infty} \left( \tan\frac{\uppi x}{1 + 2x} \right)^{\frac{1}{x}}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0^{+}} \left( \frac{\sin x}{x} \right)^{\frac{1}{1 - \cos x}}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0}\left( \frac{\cos x}{\cos 2x} \right)^{\frac{1}{x^{2}}}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0} \frac{\sin x - x\cos x}{x - \sin x}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0}\frac{1 + \frac{1}{2}x^{2} - \sqrt{1 + x^{2}}}{\left( \cos x - \ee^{\frac{x^{2}}{2}} \right) \sin \frac{x^{2}}{2}}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to \infty} \left( \sqrt[6]{x^{6} + x^{5}} - \sqrt[6]{x^{6} - x^{5}} \right)$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to +\infty}\left[ \left( x^{3} + \frac{x}{2} - \tan \frac{1}{x} \right) \ee^{\frac{1}{x}} - \sqrt{1 + x^{6}} \right]$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0}\frac{\ee^{\tan x} - \ee^{\sin x}}{x \sin^{2} x}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0} \frac{\sin x + x^{2} \sin\frac{1}{x}}{(1 + \cos x)\ln(1 + x)}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0}\left[ \frac{a}{x} - \left( \frac{1}{x^{2}} - a^{2} \right) \ln(1 + ax) \right]$，其中 $a \ne 0$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0} \frac{(1 + x)^{\frac{1}{x}} - (1 + 2x)^{\frac{1}{2x}}}{\sin x}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0}\frac{\int_{0}^{\sin^{2}x} \ln(1 + t)\dd{t}}{\left( \sqrt[3]{1 + x^{3}} - 1 \right)\sin x}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0} \frac{ \int_{0}^{x} \left[ \int_{0}^{u^{2}} \arctan(1 + t) \dd{t} \right] \dd{u} }{x(1 - \cos x)}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0^{+}} \frac{x^{x} - ( \sin x )^{x}}{x^{2}\ln(1 + x)}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0} \frac{ \cos x - \ee^{-\frac{x^{2}}{2}} }{x^{2} [ x + \ln(1 - x) ]}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0} \frac{1}{x^{3}} \left[ \left( \frac{2 + \cos x}{3} \right)^{x} - 1 \right]$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0} \frac{\ln\left( \sin^{2}x + \ee^{x} \right) - x}{\ln\left( x^{2} + \ee^{2x} \right) - 2x}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 1} \frac{x - x^{x}}{1 - x + \ln x}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0} \left( \frac{a_{1}^{x} + a_{2}^{x} + \cdots + a_{n}^{x}}{n} \right)^{\frac{1}{x}}$，$a_{i} > 0$，且 $a_{i} \ne 1, i = 1,2,\cdots,n,n \geq 2$.
	\end{ti}

	\begin{ti}
		设 $\lim_{x \to 0} \frac{\ln\left[ 1 + \frac{f(x)}{\sin x} \right]}{a^{x} - 1} = A (a > 0, a \ne 1)$，求 $\lim_{x \to 0}\frac{f(x)}{x^{2}}$.
	\end{ti}

	\begin{ti}
		已知 $\lim_{x \to 1} f(x)$ 存在，且 $f(x) = \frac{x - \arctan(x - 1) - 1}{(x - 1)^{3}} + 2x^{2} \ee^{x-1} \cdot \lim_{x \to 1} f(x)$，求 $f(x)$.
	\end{ti}

	\begin{ti}
		设函数 $f(x) = (1 + x)^{\frac{1}{x}}(x > 0)$，证明：存在常数 $A,B$，使得当 $x \to 0^{+}$ 时，恒有
		\begin{equation*}
			f(x) = \ee + Ax +Bx^{2} + o\left( x^{2} \right),
		\end{equation*}
		并求常数 $A,B$.
	\end{ti}

	\begin{ti}
		已知
		\[
			\lim_{x \to 0} \frac{(1+x)^{\frac{1}{x}} - \left( A + Bx + Cx^{2} \right)}{x^{3}} = D \ne 0.
		\]
		求常数 $A,B,C,D$.
	\end{ti}

	\begin{ti}
		设函数
		\begin{align*}
			f(x) &= \begin{cases}
				\frac{\ln\left( 1 + x^{3} \right)}{\arcsin x - x}, & x < 0,\\
				\frac{\ee^{-x} + \frac{1}{2}x^{2} + x - 1}{x \sin \frac{x}{6}}, & x > 0,
			\end{cases},\\
			g(x) &= \frac{\ee^{\frac{1}{x}}\arctan\frac{1}{x}}{1 + \ee^{\frac{2}{x}}},
		\end{align*}
		求 $\lim_{x \to 0} f[g(x)]$.
	\end{ti}

	\begin{ti}
		设 $\alpha \geq 5$ 且为常数，则 $k$ 为何值时极限
		\begin{equation*}
			I = \lim_{x \to +\infty} \left[ \left( x^{\alpha} + 8x^{4} + 2 \right)^{k} - x \right]
		\end{equation*}
		存在，并求此极限值.
	\end{ti}
	
	\begin{ti}
		已知极限
		\[
			I = \lim_{x \to 0} \left( \frac{a}{x^{2}} + \frac{b}{x^{4}} + \frac{c}{x^{5}} \int_{0}^{x} \ee^{-t^{2}} \dd{t} \right) = 1,
		\]
		求常数 $a,b,c$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 0} \frac{ \sqrt{\cos x} - \sqrt[3]{\cos x} }{\sin^{2}x}$.
	\end{ti}

	\begin{ti}
		求 $\lim_{x \to 1} \frac{\left( 1 - \sqrt[3]{x} \right) \left( 1 - \sqrt[4]{x} \right) \cdots \left( 1 - \sqrt[n]{x} \right) }{(1 - x)^{n-2}}$.
	\end{ti}
	
	\begin{ti}
		求 $\lim_{x \to 0} \frac{1 - \cos x \cdot \sqrt{\cos 2x} \cdot \sqrt[3]{\cos 3x}}{x^{2}}$.
	\end{ti}

	\begin{ti}
		设函数 $f(x)$ 满足 $f(1) = 1$，且有 $f'(x) = \frac{1}{x^{2} + f^{2}(x)}$，证明：极限 $\lim_{x \to \infty} f(x)$ 存在，且极限值小于 $1 + \frac{\uppi}{4}$.
	\end{ti}

	\begin{ti}
		设 $x \geq 0$ 时，$f(x)$ 满足 $f'(x) = \frac{1}{x^{2} + f^{2}(x)}$，且 $f(0) = 1$，证明：$\lim_{x \to +\infty} f(x)$ 存在且极限值小于 $1 + \frac{\uppi}{2}$.
	\end{ti}